Determine Target Voltage for Pre-Wired LED

Look closely at the heatshrink tubing on the LED leads: you can see a resistor-shaped bulge. Try removing the heatshrink on one and repeating your diode measurement without the resistor in circuit. Your meter in diode mode can't source enough current to drive the LED with the resistor in series. In other words, the meter won't be providing the higher voltage that was expected when the resistor was added. (This sort of pre-wired LED is commonly sold for hobbyists who just want to use them, as you said, in 5 or 12 volt situations.)

The other questions linked in my comment to your question are about calculating resistor values for LEDs to achieve a specific current. You should be familiar with those to know the basics of how LEDs aren't linear, and how the LED's forward voltage helps determine the value of resistor to use for a given desired current and a given supply voltage.

Working from your explanation, you said that you successfully lit the LED using both 5 and 12 volts. This is not surprising, but the current through the LED will be of course higher with 12 volts applied. The question then is, did the manufacturer intend the LED to be compatible with either? It could be that the current limiting resistor was chosen so that the LED works just fine at 12 V and is still suitably bright at 5 V.

You didn't mention what color the LED is but let's assume it has a forward voltage of 3 V. Let's further assume that the LED can tolerate 20 mA.

How much voltage will the resistor be "dropping" with 12 V applied? \$12 - 3 = 9 V\$

How much resistance is needed to limit current to 20 mA? \$9 / 0.02 = 450 Ω\$

How much current will the LED still see with 5 V applied? \$2 / 450 = 4.4mA\$

Most modern, efficient 5mm LEDs like the one pictured will be nice and bright even at ~4 mA.

So I'm curious, what resistor value is hiding under the heatshrink? If it's meant to provide 20 mA at 5 V, then applying 12 V will probably burn the LED up if left too long. Perhaps it's a higher resistance meant to allow it to work to voltages even higher than 12. You should do some more experimenting to find out.

(Another thing you can do if you don't want to disassemble anything, since you have a multimeter, is measure current through the pre-wired assembly and vary the supply voltage. If you assume the LED can only tolerate 20 mA at 100 % duty cycle, then whatever voltage you reach which results in 20 mA can be treated as the maximum.)

You can try to measure the resistor. One resistor end already goes to white wire, so you only have to expose only a tiny amount of the other resistor terminal under the heat shrink tubing, which might be possible without hassle if you can use one small strand from a copper wire to connect the multimeter to it. When you know which two different resistances are used, the higher one is for 12V. If you can't access the resistor otherwise, remove the tubing. Then you know the resistances when you find two different values.

If you know what you bought and have the rated current at the rated voltage, then you can measure if the rated current is achieved at 5V, if not, then it is a 12V one.

Easy.

At 5V either will work. Those wired for 12V will be less bright. This may or may not be important for you.

At 12V, either will work as well, but those wired for 5V will be way too bright and may overheat after a very short while.

Use an ampere meter at 200mA range. If the whole thing consumes less than 20mA for 5mm LED or less than 10mA for 3mm LED, you are OK.

If you don't beleive you are quick enough, test them at 5V first. If the current is about 5 times less than the values above, use at 12V.